How Is $\delta = \sqrt{9+\epsilon}-3$ the Largest Choice in a Limit Proof?

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Discussion Overview

The discussion centers on verifying the largest possible choice of $\delta$ in the context of a limit proof, specifically for the limit $\lim_{{x}\to{3}}x^2=9$. Participants explore geometric arguments and mathematical reasoning related to this limit, including the relationship between $\delta$ and $\epsilon$.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant expresses uncertainty about how to approach the problem and requests hints.
  • Another participant suggests starting with the equation $f(3)+\varepsilon=f(3+\delta)$, indicating a potential direction for the proof.
  • A different participant expands the function $f(x)=x^2$ to $f(3+\delta)=9+6\delta +(\delta)^2$, but struggles to connect this to a geometric argument.
  • A later reply provides a geometric construction involving the curve $f(x)=x^2$, describing how to visualize the relationship between $\delta$, $\epsilon$, and the limit, ultimately leading to the equation $9+\varepsilon=(3+\delta)^2$.
  • This reply suggests using the square root property to derive $\delta = \sqrt{9+\epsilon}-3$, framing it as a necessary conclusion from the geometric argument presented.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the geometric argument's clarity or effectiveness, and there is ongoing exploration of how to best illustrate the relationship between $\delta$ and $\epsilon$.

Contextual Notes

The discussion includes various assumptions about the geometric interpretation of limits and the properties of functions, which may not be universally accepted or fully resolved.

Dethrone
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Verify, by a geometric argument, that the largest possible choice of $\delta$ for showing that $\lim_{{x}\to{3}}x^2=9$ is $\delta = \sqrt{9+\epsilon}-3$

I have no clue, hints?
 
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With $f(x)=x^2$, I would begin with:

$$f(3)+\varepsilon=f(3+\delta)$$
 
I can expand it...
$f(3+\delta)=9+6\delta +(\delta)^2$, but I still don't see a geometric argument from that :(
 
Draw the curve $f(x)=x^2$. Now, on the $x$-axis at $x=3$, draw a vertical line segment up to the curve. Where this line touches the curve, draw a horizontal line segment to the left, until it reaches the $y$-axis...this is $f(3)$. Now, go back to the vertical line, and choose a distance $\delta$ from $x=3$, which will be $3+\delta$. Draw a line up to the curve, and the to the left to the $y$-axis. It will be at $f(3)+\varepsilon$. Thus we see from this construction, that we must have:

$$f(3)+\varepsilon=f(3+\delta)$$

$$9+\varepsilon=(3+\delta)^2$$

Instead of expanding, use the square root property, and take the positive root, and you will have shown what is required. :D
 

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